something like "group by" for columns?
I have table like this:
+----+---------+---------+--------+
| id | value_x | created | amount |
+----+---------+---------+--------+
value_x is set of six strings, lets say "one", "two", "three", etc. I need to create report like this:
+--------------+-------------------------+-------------------+----------------------+
| day_of_month | "one" | "two" | [etc.] |
+--------------+-------------------------+-------------------+----------------------+
| 01-01-2011 | "sum(amount) where value_x = colum name" for this specific day |
+--------------+-------------------------+-------------------+----------------------+
Most obvious solu开发者_运维问答tion is:
SELECT SUM(amount), DATE(created) FROM `table_name` WHERE value_x=$some_variable GROUP BY DATE(created)
And loop this query six times with another value for $some_variable
in every iteration, but I'm courious if is it possible to do this in single query?
What you're asking is called a "pivot table" and is typically achieved as below. The idea is for each potential value of value_x
you either produce a 1 or 0 per row and sum 1's and 0's to get the sum for each value.
SELECT
DATE(created),
SUM(CASE WHEN value_x = 'one' THEN SUM(amount) ELSE 0 END) AS 'one',
SUM(CASE WHEN value_x = 'one' THEN SUM(amount) ELSE 0 END) AS 'two',
SUM(CASE WHEN value_x = 'one' THEN SUM(amount) ELSE 0 END) AS 'three',
etc...
FROM table_name
GROUP BY YEAR(created), MONTH(created), DAY(created)
This will come close:
SELECT
s.day_of_month
,GROUP_CONCAT(CONCAT(s.value_x,':',s.amount) ORDER BY s.value_x ASC) as output
FROM (
SELECT DATE(created) as day_of_month
,value_x
,SUM(amount) as amount
FROM table1
GROUP BY day_of_month, value_x
) s
GROUP BY s.day_of_month
You will need to read the output and look for the value_x
prior to the :
to place the items in the proper column.
The benefit of this approach over @Michael's approach is that you do not need to know the possible values of field value_x
beforehand.
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